## College Algebra (10th Edition)

The infinite geometric series converges. $S_{\infty}=6$
RECALL: (1) In the infinite geometric series: $$\sum_{k=1}^{\infty}c \cdot r^{n-1}$$ $r$ is the common ratio. (2) A geometric series converges if $|r| \lt 1$. The sum of a convergent infinite geometric series is given by the formula: $S_{\infty}=\dfrac{a_1}{1-r}$ where $r$ = common ratio $a_1$ = first term $\bf\text{Solve for r}:$ Note that when a geometric series is summation notation, the expression being raised to a power is the common ratio. Thus, the common ratio of the given series is $\dfrac{2}{3}$. Since $|\frac{2}{3}|=\frac{2}{3} \lt 1$, the series converges. $\bf\text{Find the sum:}$ The first term of the series can be evaluated by substituting $1$ for $k$: $a_1 = 3\left(\dfrac{2}{3}\right)^{1} = 3\left(\dfrac{2}{3}\right) =\dfrac{6}{3}=2$ With $a_1=2$ and $r=\dfrac{2}{3}$, solve for the sum using the formula in part (2) above to obtain: $S_{\infty} = \dfrac{a_1}{a-r} \\S_{\infty}=\dfrac{2}{1-\frac{2}{3}} \\S_{\infty}=\dfrac{2}{\frac{3}{3}-\frac{2}{3}} \\S_{\infty}=\dfrac{2}{\frac{1}{3}} \\S_{\infty}=2 \cdot \dfrac{3}{1} \\S_{\infty}=6$